Question

The maximum value of $\frac{\ln x}{x}$ is:

• (A) e
• (B) $\frac{1}{e}$
• (C) $\frac{2}{e}$
• (D) 1

Answer 1

The Correct Option is B

Explanation:
Given: Let $f(x)=\frac{\ln x}{x}$Differentiating both sides, we get:$f^{\prime }(x)=\frac{d}{dx}\left(\frac{\ln x}{x}\right)$$=\frac{\frac{1}{x}(x)-1(\ln x)}{x^2}$$\Rightarrow f^{\prime }(x)=\frac{1-\ln x}{x^2}$$f^{\prime \prime }(x)=\frac{d}{dx}(f^{\prime }(x))$$=\frac{d}{dx}\left(\frac{1-\ln x}{x^2}\right)$$=\frac{x^2\frac{d}{dx}(1-\ln x)-(1-\ln x)\frac{d}{dx}\left(x^2\right)}{{\left(x^2\right)}^2}$$=\frac{x^2(-\frac{1}{x})-(1-\ln x)(2x)}{x^4}$$\Rightarrow \frac{-x-2x+2x(\ln x)}{x^4}$$\Rightarrow \frac{x(-3+2(\ln x))}{x^4}$$\Rightarrow f^{\prime \prime }(x)=\frac{-(3-2\ln x)}{x^3}$To find the value of $x$$f^{\prime }(x)=0$$\Rightarrow \frac{1-\ln x}{x^2}=0$$\Rightarrow 1-\ln x=0$$\Rightarrow \ln x=1$$\Rightarrow x=e^1=e({\ln }_{a}b=c\Rightarrow b=a^c)$Now, at $x=e$,$f^{\prime \prime }(e)=\frac{-(3-2\ln e)}{e^3}$$=\frac{-3+2}{e^3}$$=\frac{-1}{e^3}<0$At $x=e$, maximum value of $f(x)$ obtain$\therefore f(x=e)=\frac{\ln x}{x}$$=\frac{\ln e}{e}$$=\frac{1}{e}(\because \ln e=1)$Hence, the correct option is (B).